Expand Using the Binomial Theorem (2x-5y)^4

${(2 x - 5 y)}^{4}$

Step 1

Use the binomial expansion theorem to find each term. The binomial theorem states ${(a + b)}^{n} = \sum_{k = 0}^{n} n C k \cdot (a^{n - k} b^{k})$ .

$\sum_{k = 0}^{4} \frac{4!}{(4 - k)! k!} \cdot {(2 x)}^{4 - k} \cdot {(- 5 y)}^{k}$

Step 2

Expand the summation.

$\frac{4!}{(4 - 0)! 0!} \cdot {(2 x)}^{4 - 0} \cdot {(- 5 y)}^{0} + \frac{4!}{(4 - 1)! 1!} \cdot {(2 x)}^{4 - 1} \cdot {(- 5 y)}^{1} + \frac{4!}{(4 - 2)! 2!} \cdot {(2 x)}^{4 - 2} \cdot {(- 5 y)}^{2} + \frac{4!}{(4 - 3)! 3!} \cdot {(2 x)}^{4 - 3} \cdot {(- 5 y)}^{3} + \frac{4!}{(4 - 4)! 4!} \cdot {(2 x)}^{4 - 4} \cdot {(- 5 y)}^{4}$

Step 3

Simplify the exponents for each term of the expansion.

$1 \cdot {(2 x)}^{4} \cdot {(- 5 y)}^{0} + 4 \cdot {(2 x)}^{3} \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Step 4

Simplify each term.

Multiply ${(2 x)}^{4}$ by $1$ .

${(2 x)}^{4} \cdot {(- 5 y)}^{0} + 4 \cdot {(2 x)}^{3} \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Apply the product rule to $2 x$ .

$2^{4} x^{4} \cdot {(- 5 y)}^{0} + 4 \cdot {(2 x)}^{3} \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Raise $2$ to the power of $4$ .

$16 x^{4} \cdot {(- 5 y)}^{0} + 4 \cdot {(2 x)}^{3} \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Apply the product rule to $- 5 y$ .

$16 x^{4} \cdot ({(- 5)}^{0} y^{0}) + 4 \cdot {(2 x)}^{3} \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Rewrite using the commutative property of multiplication.

$16 \cdot {(- 5)}^{0} x^{4} y^{0} + 4 \cdot {(2 x)}^{3} \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Anything raised to $0$ is $1$ .

$16 \cdot 1 x^{4} y^{0} + 4 \cdot {(2 x)}^{3} \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Multiply $16$ by $1$ .

$16 x^{4} y^{0} + 4 \cdot {(2 x)}^{3} \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Anything raised to $0$ is $1$ .

$16 x^{4} \cdot 1 + 4 \cdot {(2 x)}^{3} \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Multiply $16$ by $1$ .

$16 x^{4} + 4 \cdot {(2 x)}^{3} \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Apply the product rule to $2 x$ .

$16 x^{4} + 4 \cdot (2^{3} x^{3}) \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Raise $2$ to the power of $3$ .

$16 x^{4} + 4 \cdot (8 x^{3}) \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Multiply $8$ by $4$ .

$16 x^{4} + 32 x^{3} \cdot {(- 5 y)}^{1} + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Simplify.

$16 x^{4} + 32 x^{3} \cdot (- 5 y) + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Rewrite using the commutative property of multiplication.

$16 x^{4} + 32 \cdot - 5 x^{3} y + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Multiply $32$ by $- 5$ .

$16 x^{4} - 160 x^{3} y + 6 \cdot {(2 x)}^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Apply the product rule to $2 x$ .

$16 x^{4} - 160 x^{3} y + 6 \cdot (2^{2} x^{2}) \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Raise $2$ to the power of $2$ .

$16 x^{4} - 160 x^{3} y + 6 \cdot (4 x^{2}) \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Multiply $4$ by $6$ .

$16 x^{4} - 160 x^{3} y + 24 x^{2} \cdot {(- 5 y)}^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Apply the product rule to $- 5 y$ .

$16 x^{4} - 160 x^{3} y + 24 x^{2} \cdot ({(- 5)}^{2} y^{2}) + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Rewrite using the commutative property of multiplication.

$16 x^{4} - 160 x^{3} y + 24 \cdot {(- 5)}^{2} x^{2} y^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Raise $- 5$ to the power of $2$ .

$16 x^{4} - 160 x^{3} y + 24 \cdot 25 x^{2} y^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Multiply $24$ by $25$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} + 4 \cdot {(2 x)}^{1} \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Simplify.

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} + 4 \cdot (2 x) \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Multiply $2$ by $4$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} + 8 x \cdot {(- 5 y)}^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Apply the product rule to $- 5 y$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} + 8 x \cdot ({(- 5)}^{3} y^{3}) + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Rewrite using the commutative property of multiplication.

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} + 8 \cdot {(- 5)}^{3} x y^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Raise $- 5$ to the power of $3$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} + 8 \cdot - 125 x y^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Multiply $8$ by $- 125$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} - 1000 x y^{3} + 1 \cdot {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Multiply ${(2 x)}^{0}$ by $1$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} - 1000 x y^{3} + {(2 x)}^{0} \cdot {(- 5 y)}^{4}$

Apply the product rule to $2 x$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} - 1000 x y^{3} + 2^{0} x^{0} \cdot {(- 5 y)}^{4}$

Anything raised to $0$ is $1$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} - 1000 x y^{3} + 1 x^{0} \cdot {(- 5 y)}^{4}$

Multiply $x^{0}$ by $1$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} - 1000 x y^{3} + x^{0} \cdot {(- 5 y)}^{4}$

Anything raised to $0$ is $1$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} - 1000 x y^{3} + 1 \cdot {(- 5 y)}^{4}$

Multiply ${(- 5 y)}^{4}$ by $1$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} - 1000 x y^{3} + {(- 5 y)}^{4}$

Apply the product rule to $- 5 y$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} - 1000 x y^{3} + {(- 5)}^{4} y^{4}$

Raise $- 5$ to the power of $4$ .

$16 x^{4} - 160 x^{3} y + 600 x^{2} y^{2} - 1000 x y^{3} + 625 y^{4}$

Expand Using the Binomial Theorem (2x-5y)^4

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